Chapter 5: Exercises with Solutions

Section 5.5MotionNow use either the quadratic formula or the ZERO routine on your calculator to findthe approximate solutions of this equation:x = −b ± √ b 2 − 4ac2a= −64 ± √ 64 2 − 4(−16)(−61)2(−16)= −64 ± √ 192−32≈ 1.567, 2.433Thus, it takes approximately 1.567 seconds for the rock to rise to a height of 61 feet.It also attains that height a second time on the way back down at approximately 2.433seconds. Rounded to the nearest hundredth of a second, the answer is 1.57.49. If the y-axis is oriented with the positive direction upward, and the 0 mark is setat ground level, then the height (in feet) of the water balloon after t seconds is givenby the functionNote thath(t) = −16t 2 − 24t + 169height = 0 =⇒ h(t) = 0 =⇒ −16t 2 − 24t + 169 = 0Now use either the quadratic formula or the ZERO routine on your calculator to findthe approximate solutions of this equation:x = −b ± √ b 2 − 4ac2a= 24 ± √ (−24) 2 − 4(−16)(169)2(−16)= 24 ± √ 11392−32≈ −4.085, 2.585Since −4.085 is meaningless in the context of the question, the approximate answer is2.585. Rounded to the nearest hundredth of a second, the answer is 2.59.51. If the y-axis is oriented with the positive direction upward, and the 0 mark isset at ground level, then the height (in feet) of the ball after t seconds is given by thefunctionNote thath(t) = −16t 2 + 63t + 42Version: Fall 2007